MadeEasy Test Series 2017: Operating System - Process Schedule

Question

As Operating system uses round robin process scheduling algorithm. Consider the execution times for the following processes.

$\begin{array}{|c|cl|}\hline  \text{Process Id} & \text{Brust Time}  \\ \hline A & 3\\B & 5\\C&2 \\ D&1 \\E&3\\F&4 \\\hline \end{array}$

Assume all processes arrive at time $t=0$ and cut off time =$2$ units. One cycle includes the one occurrence of each and every remain process in the system and provided that they may complete their execution in one cycle only.

How many jobs will complete their execution by the end of second cycle of selection?

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Iam confused in part that says "provided that they may complete their execution in one cycle only". If this statement is neglected then definitely answer is $5$ ( simple RR algorithm)

Answer 1

Jobs C & D require only one cycle and by the end of second cycle processes will have 4 units time(at max) hence Jobs A,E & F will have completed.

Hence , Ans is 5 i.e A,C,D,E & F.

"provided that they may complete their execution in one cycle only" simply means C & D do not require 2nd cycle (or just one cycle is enough for their execution.)

Comments

yes  in first execution cycle  only C and D complete their execution

status of each process after second execution cycle us(0,1,0,0,0,0) so 5 is right answer